1. Physics

2. Session Fluid Mechanics - 3

3. Session Objectives

4. Session Objective Viscosity & Coeff. of Viscosity Flow through narrow tube - Poiseullie's equation Stroke's law- Terminal velocity Critical velocity-Reynold's number Surface Tension Excess pressure inside a drop and soap bubble Angle of Contact Rise of liquid in a capillary tube Surface Energy

5. Viscosity The property of a fluid due to which it opposes the relative motion between the different layers is called viscosity or fluid friction. x=vt BCEF A D v F

6. Coefficient of Viscosity SI unit of viscosity is N-s/m 2 Commonly used unit is poise 1 poise = 0.1 N-s/m 2

7. Poiseuille’s Formula Poiseuille’s derived a formula for the rate of flow of a viscous fluid through a cylindrical tube which is given by whereL = Length of tube r = Radius of tube p = Pressure difference  = Viscosity of liquid

8. Terminal Velocity F W Stoke’s Law: R At a certain stand when viscous force will balance net downward force the body will fall with constant velocity called terminal velocity.

9. Critical velocity - Reynold's number The maximum velocity up to which fluid motion is steady is called critical velocity. Reynold,s through experiments established that in case of motion of fluid in thin tubes, critical velocity depends on the density (r), viscosity (h) of the fluid and radius r of the tube.

10. Surface Tension AB F F If AB = L SI unit of Surface tension:Nm -1 The property of a liquid due to which its free surface tries to have minimum surface area and behaves as if it were under tension somewhat like a stretched elastic membrane, is called surface tension.

11. Surface Energy Surface energy = surface tension x surface area U = T X A A molecule in the surface has greater potential energy than a molecule well inside the liquid. The extra energy that a surface layer has is called the surface energy

12. Excess pressure inside a drop Excess pressure inside a soap bubble P1P1 P 2 -P 1 =4T/R P2P2 P2P2 P 2 -P 1 =2T/R P1P1 Excess pressure inside a drop and drop bubble

13. Angle of Contact  “Angle of contact is defined as the angle between the tangent to solid and liquid surfaces at a point of contact inside the liquid.” It depends upon the nature of solid and liquid. For concave miniscus it is acute while for convex it is obstuse. 

14. Capillary Rise  Is the angle between the tangents to the solid and liquid surface at the point of contact.(angle of contact) Where r is radius of capillary and R is the radius of meniscus r   h P0P0 R

15. Class Test

16. Class Exercise - 1 The energy needed in breaking a drop of radius R into n drops of radius r is (a) (b) (c) (d)

17. Solution Hence, answer is (a).

18. Class Exercise - 2 A very narrow capillary tube records a rise of 20 cm when dipped in water. When the area of cross section is reduced to one-forth of former value, water will rise a height of (a) 10 cm(b) 20 cm (c) 40 cm(d) 80 cm

19. Solution Area reduced to one-forth, i.e. radius reduced to half h 2 = 2h 1 h 2 = 2 x 20 = 40 cm Hence, answer is (c).

20. Class Exercise - 3 The height of a liquid column in the capillary on the surface of the moon. If h is on the surface of the earth. (a) h(b) (c) 6h(d) zero

21. Solution Hence, answer is (c).

22. Class Exercise - 4 The viscous force on a small sphere of radius R moving in a fluid varies as (a) R 2 (b) R (c) (d)

23. Solution From Stokes’s equation, Hence, answer is (b).

24. Class Exercise - 5 A solid sphere falls with a terminal velocity of 10 cm/s in earth’s gravitational field. If it is allowed to full in a region outside the gravitational field of earth, the terminal velocity will be (a) equal to 10 cm/s (b) more than 10 cm/s (c) less than 10 cm/s (d) zero

25. Solution In gravity-free region g = 0 Hence, answer is (d).

26. Class Exercise - 6 Two soap bobbles with radii 3 cm and 4 cm coolesce in vacuum under isothermal conditions to form a bigger bobble of radius R. Then R is equal to (a) (b) (c) 5 cm(d) 7 cm

27. Solution Since the soap bobbles coalese in vacuum under isothermal condition, there is no change in free surface area. R 2 = 3 2 + 4 2 R 2 = 25 R = 5 cm Hence, answer is (c).

28. Class Exercise - 7 If more air is pushed into a soap bobble, the pressure inside it (a) increases (b) decreases (c) remains the same (d) becomes infinite

29. Solution Pressure inside soap bobble = P 0 – atmospheric pressure As more air is pushed into it r increases. So decreases as a whole pressure decreases. Hence, answer is (b).

30. Class Exercise - 8 Drops of a liquid of density D and surface tension T are floating in liquid of density d with half of drop in the liquid. Then the radius of the drop is (a) (b) (c) (d)

31. Solution Balancing forces Hence, answer is (b).

32. Class Exercise - 9 A soap bobble has radius r and volume V. If the excess pressure inside the boubble is P, then PV is proportional to (a) r 4 (b) r 3 (c) r 2 (d) r

33. Solution Hence, answer is (c).

34. Thank you